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myrzilka [38]
3 years ago
15

How many grams water will condense when 56,500 joules of energy is removed from steam at its boiling point

Chemistry
1 answer:
irina1246 [14]3 years ago
5 0

Answer:

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I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity o
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We know that molarity = mol/L, so:

1.5 M = x mol/0.345 L

1.5 M * 0.345 L = x mol => 0.5175 mol

0.5175 mol/0.250 L = 2.07 M

Your new molarity of the solution will be 2.07 M.
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What else supports the idea that an exothermic reaction has occurred?
kodGreya [7K]
The main pieces of evidence that an exothermic reaction has occurred is an increase in temperature due to the release of energy, a release of energy in the form of light, or a release of gas.
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NEED HELP ASAP WILL GIVE BRAINLYEST Organisms that are made up of prokaryotic cells are NOT ____________. Multicellular All of t
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Explanation:

the answer is that prokaryotic cells are not multicellular

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3 years ago
There are many positive and negative aspects associated with nuclear power. What is a positive aspect associated with nuclear po
Keith_Richards [23]

Answer: A)There is a high ratio of energy to cost

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A Chunk of Uranium can power a city a lot longer than a chunk of coal. 
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It is very expensive to build and maintain a nuclear power plant at first so investors whom want money up front are more reluctant to loan money for one.

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5 0
3 years ago
Read 2 more answers
Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
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